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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 11, Problem 53E
Textbook Problem
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Consider the following initial rate data for the decomposition of compound AB to give A and B:

[AB]0 (mol/L) Initial Rate (mol/L · s|)
0.200 3.20 × 10−3
0.400 1.28 × 10−2
0.600 2.88 × 10−2

Determine the half-life for the decomposition reaction initially having 1.00 M AB present.

Interpretation Introduction

Interpretation: The data analysis table and initial concentration of reactant AB is given. By using these values, the half life for decomposition reaction is to be calculated.

Concept introduction: The half-life is defined as the time required for the concentration of reactant to be reduced to one-half of its initial value.

The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction.

To determine: The half life for the decomposition reaction initially having 1.00M AB .

Explanation of Solution

Given

The initial concentration of AB is 1.00M .

The reaction of decomposition of AB is,

ABA+B

The rate law gives the relation between reaction rate and concentration of reactants. The rate law is represented as,

Rate=k[AB]a (1)

Where,

  • k is rate constant.
  • [AB] is concentration of reactant.
  • a is reaction order.

The value of a is calculated by comparing the different rates from the given table.

The given table is,

[AB]0 (mol/L) Initial rate (mol/Ls)
0.200 3.20×103
0.400 1.28×102
0.600 2.88×102

For the first row:

Substitute the value of rate and concentration of AB in equation (1).

3.20×103mol/Ls=k(0.200)a (2)

For the second raw:

Substitute the value of rate and concentration of AB in equation (1).

1.28×102mol/Ls=k(0.400)a (3)

For the third raw:

Substitute the value of rate and concentration of AB in equation (1).

2.88×102mol/Ls=k(0.600)a (4)

Divide equation (3) by equation (1).

1.28×102mol/Ls3.20×103mol/Ls=k(0

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Chapter 11 Solutions

Chemistry: An Atoms First Approach
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